Question

Part 3: Perturbation: We want to get the perturbation Hamiltonian H1which includes the first correction to KE Hamiltonian Ho that, we expand (wrt small P) the relativistic KE (in terms of momentum operators), from special relativity. To do P2 C2+m2c4 - mc T E-mc2 = +.. Но The first term inside the dots . .- is Hwhich is A. рз НH 2m В. рз Нi 4m2c С. ра Н 2m D. pа 8m32 Н

Answer 1

The correct option will be  D.  we can calculate this by expanding the expression by binomial theorem.

- mc2 - 1p4 We have Ta MEA +p2c2 1. 1 = mc2/17 8 much 2n T= lm - mc2 - -me2 me/ m² + p2 - 4 = mc² j1 + P2 _mc2 m2,2 1/2 mc² 2m2 8m3 (2 -me wK Ha HotH 2 = mc² ema
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